2012 ပထမလပတ္ စာေမးပြဲ ေမးခြန္းနဲ႔ အေျဖ

မေရးတာ ၾကာၿပီ။ အခုျပန္ေရးပါတယ္။ အဓိကေတာ့ ေလ့က်င့္ခန္းေတြပဲ ဦးစားေပး တင္ပါေတာ့မယ္ အဆင္ေျပညီညြတ္တဲ့ အခါလည္း သခၤန္းစာေတြ ေရးမွာပါ။ အခု 2012 ပထမဦးဆံုး လပတ္ စာေမးပြဲ ေမးခြန္းနဲ႔ အေျဖကို တင္ေပးလိုက္ပါတယ္။ ေအာက္မွာေပးထားတဲ့ download link ေတြကေန ဆြဲယူႏိုင္ပါတယ္။

Grade (11)



Grade (10)

Factor Formulae - Derivation

$ \displaystyle \begin{array}{l}\sin (\alpha +\beta )=\sin \alpha \cos \beta +\cos \alpha \sin \beta ---(1)\\\\ \sin (\alpha -\beta )=\sin \alpha \cos \beta -\cos \alpha \sin \beta ---(2)\ \ \ \end{array}$

ဆိုတဲ့ Sum and Difference Formulae ေတြ ကို မွတ္မိၾကမယ္ ထင္ပါတယ္။

ညီမွ်ျခင္း (1) နဲ႔ (2) ကို ေပါင္းလိုက္မယ္...။

$ \displaystyle \sin (\alpha +\beta )+\ \sin (\alpha -\beta )=2\sin \alpha \cos \beta ---(*)$

ဆိုၿပီး ရလာမွာေပါ့...။

$ \displaystyle \alpha +\beta =\theta $ နဲ႔ $ \displaystyle \alpha -\beta =\phi $ လို႔ ထားလိုက္မယ္။

ဒါဆိုရင္ $ \displaystyle \alpha =\frac{{\theta +\phi }}{2}$ နဲ႔ $ \displaystyle \beta =\frac{{\theta -\phi }}{2}$ ျဖစ္သြားမွာေပါ့...။

$ \displaystyle \alpha +\beta =\theta ,\ \alpha -\beta =\phi ,\ \ \alpha =\frac{{\theta +\phi }}{2},\beta =\frac{{\theta -\phi }}{2}$ တို႔ကို (*) မွာ အစားသြင္းလိုက္တဲ့ အခါ မွာေတာ့ ေအာက္ပါ factor formula ကို ရရွိမွာ ျဖစ္ပါတယ္။

$ \displaystyle \sin \theta +\ \sin \phi =2\sin \frac{{\theta +\phi }}{2}\cos \frac{{\theta -\phi }}{2}$

ဒီတစ္ခါ ညီမွ်ျခင္း (1) ထဲက (2) ကို ႏႈတ္ပါမယ္...။

$ \displaystyle \sin (\alpha +\beta )-\ \sin (\alpha -\beta )=2\cos \alpha \sin \beta $

အထက္ကအတိုင္း သက္ဆိုင္ရာတန္ဖိုးေတြ အစားသြင္းလိုက္ရင္ ...။

$ \displaystyle \sin \theta -\ \sin \phi =2\cos \frac{{\theta +\phi }}{2}\sin \frac{{\theta -\phi }}{2}$

Identity တစ္ခု ထပ္ရပါမယ္...။

ေနာက္ထပ္ညီမွ်ျခင္း ႏွစ္ၾကာင္း ကို ဆက္ၾကည့္ရေအာင္...။

$ \displaystyle \begin{array}{l} \cos (\alpha +\beta )=\cos \alpha \cos \beta -\sin \alpha \sin \beta ---(3)\\\\ \cos (\alpha -\beta )=\cos \alpha \cos \beta +\sin \alpha \sin \beta ---(4)\ \ \ \end{array}$

အထက္ပါအတိုင္း ညီမ ွ်ျခင္း ႏွစ္ေၾကာင္း (3) နဲ႔ (4) ကို ေပါင္းတစ္လွည့္ ႏႈတ္တစ္လွည့္ လုပ္လိုက္ရင္ ...။

$ \displaystyle \begin{array}{l}(3)+(4)\Rightarrow \ \ \ \cos (\alpha +\beta )+\ \cos (\alpha -\beta )=2\cos \alpha \cos \beta \\\\(3)-(4)\Rightarrow \ \ \ \cos (\alpha +\beta )-\ \cos (\alpha -\beta )=-2\cos \alpha \cos \beta \end{array}$

အထက္မွာ ရွာခဲ့ၿပီး ျဖစ္တဲ့ $ \displaystyle \alpha +\beta =\theta ,\ \alpha -\beta =\phi ,\ \ \alpha =\frac{{\theta +\phi }}{2},\beta =\frac{{\theta -\phi }}{2}$ တို႔ကို သက္ဆိုင္ရာ တန္ဖိုးေတြမွာ အစားသြင္းလိုက္ရင္...။

$ \displaystyle \begin{array}{l}\cos \theta +\ \cos \phi =2\cos \displaystyle \frac{{\theta +\phi }}{2}\cos \displaystyle \frac{{\theta -\phi }}{2}\\\\\cos \theta -\ \cos \phi =-2\sin \displaystyle \frac{{\theta +\phi }}{2}\sin \displaystyle \frac{{\theta -\phi }}{2}\end{array}$

Half-Angle Formulae - Derivation

$ \displaystyle \ \cos 2\alpha =1-2{{\sin }^{2}}\alpha $ ဆိုတာ သိခဲ့ပါၿပီ။

$ \displaystyle 2\alpha = \theta$ လို႔ ထားလိုက္မယ္။ ဒါဆိုရင္ $ \displaystyle \theta =\frac{\alpha}{2}$ ေပါ့...။

မူလညီမွ်ျခင္းမွာ အစားသြင္းလိုက္ ရင္ ...

$ \displaystyle \begin{array}{l}2{{\sin }^{2}}\displaystyle \frac{\theta }{2}=1-\cos \theta \\\\{{\sin }^{2}}\displaystyle \frac{\theta }{2}=\displaystyle \frac{{1-\cos \theta }}{2}\end{array}$

ဒါ့ေၾကာင့္ . . .

$ \displaystyle \sin \frac{\theta }{2}=\pm \sqrt{{\frac{{1-\cos \theta }}{2}}}$

$ \displaystyle \cos 2\alpha =2{{\cos }^{2}}\alpha -1$ လို႔လည္း သိထားခဲ့ၿပီးသား မဟုတ္လား . . .။

အထက္မွာ ေျပာခဲ့တဲ့အတိုင္း $ \displaystyle 2\alpha =\theta ,\alpha =\frac{\theta }{2}$ ကို အစားသြင္းလိုက္ရင္ ...

$ \displaystyle \begin{array}{l}2{{\cos }^{2}}\displaystyle \frac{\theta }{2}=1+\cos \theta \\\\{{\cos }^{2}}\displaystyle \frac{\theta }{2}=\displaystyle \frac{{1+\cos \theta }}{2}\end{array}$

ဒါ့ေၾကာင့္ . . .

$ \displaystyle \cos \frac{\theta }{2}=\pm \sqrt{{\frac{{1+\cos \theta }}{2}}}$

$ \displaystyle \sin \frac{\theta }{2}$ နဲ႕ $ \displaystyle \cos \frac{\theta }{2}$ ကို သိၿပီဆိုေတာ့ ....

$ \displaystyle \tan \displaystyle \frac{\theta }{2}= \displaystyle \frac{{\sin \displaystyle \frac{\theta }{2}}}{{\cos \displaystyle \frac{\theta }{2}}}$ ဆိုတဲ့ basic identity ကို သံုးၿပီး ဆက္ရွာလို႔ရၿပီေပါ့။

$ \displaystyle \begin{array}{*{20}{l}} {\tan \displaystyle \frac{\theta }{2}=\displaystyle \frac{{\sin \displaystyle \frac{\theta }{2}}}{{\cos \displaystyle \frac{\theta }{2}}}} \\ {} \\ {\tan \displaystyle \frac{\theta }{2}=\pm \displaystyle \frac{{\sqrt{{\displaystyle \frac{{1-\cos \theta }}{2}}}}}{{\sqrt{{\displaystyle \frac{{1+\cos \theta }}{2}}}}}} \\ {} \\ {\tan \displaystyle \frac{\theta }{2}=\pm \sqrt{{\displaystyle \frac{{\displaystyle \frac{{1-\cos \theta }}{2}}}{{\displaystyle \frac{{1+\cos \theta }}{2}}}}}} \end{array}$

ဒါ့ေၾကာင့္ . . .

$ \displaystyle \tan \displaystyle \frac{\theta }{2}=\pm \sqrt{{\displaystyle \frac{{1-\cos \theta }}{{1+\cos \theta }}}}$

ဆက္ၿပီး derive လုပ္ၾကည့္မယ္ ...။

$ \displaystyle \begin{array}{*{20}{l}} {\tan \displaystyle \frac{\theta }{2}=\sqrt{{\displaystyle \frac{{1-\cos \theta }}{{1+\cos \theta }}\times \displaystyle \frac{{1+\cos \theta }}{{1+\cos \theta }}}}} \\ {} \\ {\tan \displaystyle \frac{\theta }{2}=\sqrt{{\displaystyle \frac{{1-{{{\cos }}^{2}}\theta }}{{{{{(1+\cos \theta )}}^{2}}}}}}} \\ {} \\ {\tan \displaystyle \frac{\theta }{2}=\sqrt{{\displaystyle \frac{{{{{\sin }}^{2}}\theta }}{{{{{(1+\cos \theta )}}^{2}}}}}}} \end{array}$

ဒါ့ေၾကာင့္ . . .

$ \displaystyle \tan \frac{\theta }{2}=\frac{{\sin \theta }}{{1+\cos \theta }}$

တဖန္ . . .

$ \displaystyle \begin{array}{*{20}{l}} {\tan \displaystyle \frac{\theta }{2}=\sqrt{{\displaystyle \frac{{1-\cos \theta }}{{1+\cos \theta }}\times \displaystyle \frac{{1-\cos \theta }}{{1-\cos \theta }}}}} \\ {} \\ {\tan \displaystyle \frac{\theta }{2}=\sqrt{{\displaystyle \frac{{{{{(1-\cos \theta )}}^{2}}}}{{1-{{{\cos }}^{2}}\theta }}}}} \\ {} \\ {\tan \displaystyle \frac{\theta }{2}=\sqrt{{\displaystyle \frac{{{{{(1-\cos \theta )}}^{2}}}}{{{{{\sin }}^{2}}\theta }}}}} \end{array}$

ဒါ့ေၾကာင့္ . . .

$ \displaystyle \tan \frac{\theta }{2}=\frac{{1-\cos \theta }}{{\sin \theta }}$

Double Angle Formulae - Derivation

$ \displaystyle \sin (\alpha +\beta )=\sin \alpha \cos \beta +\cos \alpha \sin \beta $ ဆိုတာ သိခဲ့ၿပီး ျဖစ္မယ္ ထင္ပါတယ္။

ဒီ ပံုေသနည္းဟာ မည္သည့္ေထာင့္ $ \displaystyle \alpha$ နဲ႔ $ \displaystyle \beta$ အတြက္မဆို မွန္ပါတယ္။

ဒါဆိုရင္ $ \displaystyle \alpha=\beta$ အတြက္လည္း မွန္တာေပါ့။... ဒါေၾကာင့္

$ \displaystyle \sin (\alpha +\beta )=\sin \alpha \cos \beta +\cos \alpha \sin \beta $

$ \displaystyle \alpha=\beta,$ ျဖစ္တဲ့အခါ

$ \displaystyle \sin (\alpha +\alpha )=\sin \alpha \cos \alpha +\cos \alpha \sin \alpha $

ဒါ့ေၾကာင့္

$ \displaystyle \sin 2\alpha =2\sin \alpha \cos \alpha $

အလားတူပါပဲ.....။

$ \displaystyle \cos (\alpha +\beta )=\cos \alpha \cos \beta -\sin \alpha \sin \beta $

$ \displaystyle \alpha=\beta,$ ျဖစ္တဲ့အခါ

$ \displaystyle \cos (\alpha +\alpha )=\cos \alpha \cos \alpha -\sin \beta \sin \beta $

ဒါ့ေၾကာင့္...

$ \displaystyle \cos 2\alpha ={{\cos }^{2}}\alpha -{{\sin }^{2}}\alpha $

$ \displaystyle {{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha =1$ ဆိုတဲ့ Pythagorean Identity ကို မွတ္မိမယ္ ထင္ပါတယ္။

$ \displaystyle {{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha =1$ ျဖစ္တာေၾကာင့္ $ \displaystyle {{\sin }^{2}}\alpha =1-{{\cos }^{2}}\alpha $ နဲ႔ $ \displaystyle {{\cos }^{2}}\alpha =1-{{\sin }^{2}}\alpha $ ျဖစ္ပါတယ္။

$ \displaystyle \cos 2\alpha ={{\cos }^{2}}\alpha -{{\sin }^{2}}\alpha $ ဆိုတဲ့ equation မွာ သက္ဆိုင္ရာ တန္ဖိုးေတြကို အစားသြင္းလိုက္ရင္ ...

$ \displaystyle \begin{array}{l}\cos 2\alpha ={{\cos }^{2}}\alpha -{{\sin }^{2}}\alpha \\\\\cos 2\alpha =1-{{\sin }^{2}}\alpha -{{\sin }^{2}}\alpha \end{array}$

ဒါ့ေၾကာင့္...

$ \displaystyle \cos 2\alpha =1-2{{\sin }^{2}}\alpha $

အလားတူပါပဲ...။

$ \displaystyle \begin{array}{l}\cos 2\alpha ={{\cos }^{2}}\alpha -{{\sin }^{2}}\alpha \\\\\cos 2\alpha ={{\cos }^{2}}\alpha -(1-{{\cos }^{2}}\alpha )\end{array}$

ဒါ့ေၾကာင့္...

$ \displaystyle \cos 2\alpha =2{{\cos }^{2}}\alpha -1$

$ \displaystyle \tan 2\alpha $ အတြက္ ဆက္ရွာၾကည့္ပါမယ္။

$ \displaystyle \tan (\alpha +\beta )=\frac{{\tan \alpha +\tan \beta }}{{1-\tan \alpha \tan \beta }}$ လို႔ သိခဲ့ၿပီးပါၿပီ။..

$ \displaystyle \alpha=\beta,$ ျဖစ္တဲ့အခါ

$ \displaystyle \tan (\alpha +\alpha )=\frac{{\tan \alpha +\tan \alpha }}{{1-\tan \alpha \tan \alpha }}$ ..

ဒါ့ေၾကာင့္...

$ \displaystyle \tan 2\alpha =\frac{{2\tan \alpha }}{{1-{{{\tan }}^{2}}\alpha }}$

Sum and Difference Formulae - Derivation

$ \displaystyle ΔABC, ΔACD$ နဲ႔ $ \displaystyle ΔCDF$ တို႔ဟာ ေထာင့္မွန္ႀတိဂံမ်ား ျဖစ္ၾကပါတယ္။

$ \displaystyle ΔABC$ မွာ $ \displaystyle ∠CAB$ ကို $ \displaystyle \alpha$ လို႔ သတ္မွတ္ပါမယ္။

$ \displaystyle ΔABC\simΔCDF$ ျဖစ္တာေၾကာင့္ $ \displaystyle ∠CDF=\alpha$ ျဖစ္ပါတယ္။

$ \displaystyle ΔACD$ မွာေတာ့ $ \displaystyle ∠CAD$ ကို $ \displaystyle \beta$ လို႔ သတ္မွတ္ပါမယ္။

ဒါဆိုရင္ $ \displaystyle ΔABC$ မွာ...

$ \displaystyle \sin \alpha=\frac{BC}{AC}$ နဲ႕ $ \displaystyle \cos \alpha=\frac{AB}{AC}$ ျဖစ္ပါတယ္။

ဒါ့ေၾကာင့္ $ \displaystyle BC =AC \sin \alpha$ နဲ႕ $ \displaystyle AB =AC \cos \alpha$ လို႔ ဆိုႏိုင္ပါတယ္။

တဖန္ $ \displaystyle ΔCDF$ မွာ...

$ \displaystyle \sin \alpha=\frac{FC}{DC}$ နဲ႕ $ \displaystyle \cos \alpha=\frac{DF}{DC}$ ျဖစ္ပါတယ္။

ဒီမွာလည္း $ \displaystyle FC =DC \sin \alpha$ နဲ႕ $ \displaystyle DF =DC \cos \alpha$ လို႔ ဆိုႏိုင္ပါတယ္။

$ \displaystyle ΔACD$ မွာလည္း ...

$ \displaystyle \sin \beta=\frac{DC}{AD}$ နဲ႕ $ \displaystyle \cos \beta=\frac{AC}{AD}$ ျဖစ္ပါတယ္။

ဒါဆိုရင္ $ \displaystyle DC =AD \sin \alpha$ နဲ႕ $ \displaystyle AC =AD \cos \alpha$ လို႔ ဆိုႏိုင္ပါတယ္။

$ \displaystyle ΔADE$ အတြက္ ဆက္ၾကည့္ရေအာင္...

$ \displaystyle \sin (\alpha+\beta)=\frac{DE}{AD}$ ျဖစ္ပါတယ္။

ပံုမွာ ေတြ႔ရတဲ့ အတိုင္း $ \displaystyle DE =DF+FE$ ျဖစ္ပါတယ္။။

ဒါ့ေၾကာင့္ $ \displaystyle \sin (\alpha +\beta )=\frac{{DF}}{{AD}}+\frac{{FE}}{{AD}}$ လို႔ ေျပာႏိုင္ပါတယ္။ ။

တဖန္ ့ $ \displaystyle BCFE$ က rectangle ျဖစ္တာေၾကာင့္ $ \displaystyle FE=BC$ လို႔ ေျပာႏိုင္ျပန္ပါတယ္။ ။

ဒါ့ေၾကာင့္ $ \displaystyle \sin (\alpha +\beta )=\frac{{DF}}{{AD}}+\frac{{BC}}{{AD}}$ လို႔ ေျပာႏိုင္ျပန္ပါတယ္။။

$ \displaystyle DF =DC \cos \alpha, BC =AC \sin \alpha$ လို႔ အထက္မွာ သိခဲ့ၿပီးပါၿပီ။ ဒါဆိုရင္ ။

$ \displaystyle \sin (\alpha +\beta )=\frac{{DC}}{{AD}}\cos \alpha +\frac{{AC}}{{AD}}\sin \alpha $ လို႔ ေျပာလို႔ရတာေပါ့။ ။

ဒီအခါမွာလည္း $ \displaystyle \sin \beta=\frac{DC}{AD}, \cos \beta=\frac{AC}{AD}$ လို႕သိခဲ့ၿပီးပါၿပီ။ ဒါေၾကာင့္။

$ \displaystyle \sin (\alpha +\beta )=\sin \beta \cos \alpha +\cos \beta \sin \alpha $ လို႔ ေျပာလို႔ရပါၿပီ။ ျပန္စီလိုက္ရင္ ...။

$ \displaystyle \sin (\alpha +\beta )=\sin \alpha \cos \beta +\cos \alpha \sin \beta $

အထက္ပါအတိုင္း ... $ \displaystyle \cos (\alpha +\beta )$ အတြက္ ပံုေသနည္းကို ဆက္ရွာႏိုင္ပါတယ္။ ..။

$ \displaystyle \begin{array}{l}\cos (\alpha +\beta )=\displaystyle \frac{{AE}}{{AD}}\\\\\cos (\alpha +\beta )=\displaystyle \frac{{AB-EB}}{{AD}}\\\\\cos (\alpha +\beta )=\displaystyle \frac{{AB-FC}}{{AD}}\ \ \ \ \left[ {\because EB=FC} \right]\\\\\cos (\alpha +\beta )=\displaystyle \frac{{AB}}{{AD}}-\displaystyle \frac{{FC}}{{AD}}\\\\\cos (\alpha +\beta )=\displaystyle \frac{{AC}}{{AD}}\cos \alpha -\displaystyle \frac{{DC}}{{AD}}\sin \alpha \\\\\text{Since}\ \displaystyle \frac{{AC}}{{AD}}=\cos \beta \ \operatorname{and}\ \displaystyle \frac{{DC}}{{AD}}=\sin \beta ,\\\\\text{Therefore,}\end{array}$

$ \displaystyle \cos (\alpha +\beta )=\cos \alpha \cos \beta -\sin \alpha \sin \beta $

$ \displaystyle \sin (\alpha +\beta )$ နဲ႔ $ \displaystyle \cos (\alpha +\beta )$ ကို သိၿပီဆိုေတာ့ $ \displaystyle \tan (\alpha +\beta )$ ကို ရွာႏိုင္ၿပီေပါ့။

$ \displaystyle \begin{array}{l}\tan (\alpha +\beta )=\displaystyle \frac{{\sin (\alpha +\beta )}}{{\cos (\alpha +\beta )}}\\\\\tan (\alpha +\beta )=\displaystyle \frac{{\sin \alpha \cos \beta +\cos \alpha \sin \beta }}{{\cos \alpha \cos \beta -\sin \alpha \sin \beta }}\\\\\text{Dividing the numerator and denominator }\\\text{with}\ \cos \alpha \cos \beta ,\\\\\tan (\alpha +\beta )=\displaystyle \frac{{\displaystyle \frac{{\sin \alpha \cos \beta }}{{\cos \alpha \cos \beta }}+\displaystyle \frac{{\cos \alpha \sin \beta }}{{\cos \alpha \cos \beta }}}}{{\displaystyle \frac{{\cos \alpha \cos \beta }}{{\cos \alpha \cos \beta }}-\displaystyle \frac{{\sin \alpha \sin \beta }}{{\cos \alpha \cos \beta }}}}\\\\\text{Therefore,}\end{array}$

$ \displaystyle \tan (\alpha +\beta )=\frac{{\tan \alpha +\tan \beta }}{{1-\tan \alpha \tan \beta }}$

$ \displaystyle \begin{array}{l}\ \ \ \ \sin (-\alpha )=-\sin \alpha ,\\\\\ \ \ \ \cos (-\alpha )=\cos \alpha ,\\\\\ \ \ \ \tan (-\alpha )=-\tan \alpha \\\\\ \ \ \ \sin \left( {\alpha -\beta } \right)=\sin \left[ {\alpha +(-\beta )} \right]\\\\\ \ \ \ \sin \left( {\alpha -\beta } \right)=\sin \alpha \cos (-\beta )+\cos \alpha \sin (-\beta )\\\\\text{Therefore,}\end{array}$

$ \displaystyle \sin \left( {\alpha -\beta } \right)=\sin \alpha \cos \beta -\cos \alpha \sin \beta $

$ \displaystyle \begin{array}{l}\ \ \ \ \cos \left( {\alpha -\beta } \right)=\cos \left[ {\alpha +(-\beta )} \right]\\\\\ \ \ \ \cos \left( {\alpha -\beta } \right)=\cos \alpha \cos (-\beta )-\sin \alpha \sin (-\beta )\\\\\text{Therefore,}\end{array}$

$ \displaystyle \cos \left( {\alpha -\beta } \right)=\cos \alpha \cos \beta +\sin \alpha \sin \beta $

$ \displaystyle \begin{array}{l}\ \ \ \ \tan \left( {\alpha -\beta } \right)=\tan \left[ {\alpha +(-\beta )} \right]\\\\\ \ \ \ \tan \left( {\alpha -\beta } \right)=\displaystyle \frac{{\tan \alpha +\tan (-\beta )}}{{1-\tan \alpha \tan (-\beta )}}\end{array}$

$ \displaystyle \tan \left( {\alpha -\beta } \right)=\frac{{\tan \alpha -\tan \beta }}{{1+\tan \alpha \tan \beta }}$

အားလံုးျပန္ေပါင္းရရင္....

$ \displaystyle \begin{array}{l} \sin \left( {\alpha \pm \beta } \right)=\sin \alpha \cos \beta \pm \cos \alpha \sin \beta \\\\ \cos \left( {\alpha \pm \beta } \right)=\cos \alpha \cos \beta \mp \sin \alpha \sin \beta \\\\ \tan \left( {\alpha \pm \beta } \right)=\displaystyle \frac{{\tan \alpha \pm \tan (-\beta )}}{{1\mp \tan \alpha \tan (-\beta )}}\end{array}$

Exercise (11.3) No(4, 5) - Solution

4. If $ \displaystyle α + β + γ = 180°$, prove that

$ \displaystyle \text{(a)}\ \ \sin (\alpha +\beta )=\cos (90{}^\circ -\gamma )$

$ \displaystyle \text{(b)}\ \ \sin (\frac{{\alpha +\beta }}{2})=\sin (90{}^\circ +\frac{\gamma }{2})$

$ \displaystyle \text{(c)}\ \ \tan \left( {\frac{\alpha }{2}} \right)=\cot \left( {180{}^\circ +\frac{{\beta +\gamma }}{2}} \right)$

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$ \displaystyle \text{(a)}\ \ \alpha +\beta +\gamma =180{}^\circ $

$ \displaystyle \begin{array}{l}\ \ \ \ \ \ \alpha +\beta =180{}^\circ -\gamma \\\\\therefore \ \ \ \ \sin (\alpha +\beta )=\sin (180{}^\circ -\gamma )\\\ \ \ \\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\sin \gamma \\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\ \cos (90{}^\circ -\gamma )\end{array}$


$ \displaystyle \begin{array}{l}\text{(b)}\ \ \alpha +\beta +\gamma =180{}^\circ \\\\\ \ \ \ \ \ \alpha +\beta =180{}^\circ -\gamma \\\\\ \ \ \ \ \ \displaystyle \frac{{\alpha +\beta }}{2}=\displaystyle \frac{{180{}^\circ -\gamma }}{2}=90{}^\circ \displaystyle -\frac{\gamma }{2}\\\\\therefore \ \ \ \ \sin (\displaystyle \frac{{\alpha +\beta }}{2})=\sin (90{}^\circ \displaystyle -\frac{\gamma }{2})\\\ \ \ \\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\cos \displaystyle \frac{\gamma }{2}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\ \sin (90{}^\circ +\displaystyle \frac{\gamma }{2})\end{array}$


$ \displaystyle \begin{array}{l}\text{(c)}\ \ \alpha +\beta +\gamma =180{}^\circ \\\\\ \ \ \ \ \ \alpha =180{}^\circ -(\beta +\gamma )\\\\\ \ \ \ \ \ \displaystyle \frac{\alpha }{2}=\displaystyle \frac{{180{}^\circ -(\beta +\gamma )}}{2}=90{}^\circ \displaystyle -\frac{{\beta +\gamma }}{2}\\\\\therefore \ \ \ \ \tan \left( {\displaystyle \frac{\alpha }{2}} \right)=\tan (90{}^\circ \displaystyle -\frac{{\beta +\gamma }}{2})\\\ \ \ \\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\cot \displaystyle \frac{{\beta +\gamma }}{2}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\ \cot \left( {180{}^\circ +\displaystyle \frac{{\beta +\gamma }}{2}} \right)\end{array}$

5.    Prove that in any triangle $ \displaystyle ABC,$

(i) $ \displaystyle \sin (A+B) = \sin C.$

(ii) $ \displaystyle \cos(A+B) + \cos C = 0.$

(iii) $ \displaystyle \cos \frac{A+B}{2} = \sin \frac{C}{2}.$

(iv) $ \displaystyle \tan \frac{A+B}{2} = \cot \frac{C}{2}.$

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(i) $ \displaystyle \text{Since}\ A+B+C=180{}^\circ ,$

$ \displaystyle \begin{array}{l}\therefore A+B=180{}^\circ -C\\\\\therefore \sin (A+B)=\sin (180{}^\circ -C)\\\\\therefore \sin (A+B)=\sin C\end{array}$


(ii) $ \displaystyle \text{Similarly, }\cos (A+B)=\cos (180{}^\circ -C)$

$ \displaystyle \begin{array}{l}\therefore \cos (A+B)=-\cos C\\\\\therefore \cos (A+B)+\cos C=0\end{array}$


(iii) $ \displaystyle \cos \left( {\frac{{A+B}}{2}} \right)=\cos \left( {\frac{{180{}^\circ -C}}{2}} \right)$

$ \displaystyle \therefore \ \cos \left( {\frac{{A+B}}{2}} \right)=\cos \left( {90{}^\circ -\frac{C}{2}} \right)$

$ \displaystyle \therefore \ \cos \left( {\frac{{A+B}}{2}} \right)=\sin \frac{C}{2}$


(iv) $ \displaystyle \tan \left( {\frac{{A+B}}{2}} \right)=\tan \left( {\frac{{180{}^\circ -C}}{2}} \right)$

$ \displaystyle \therefore \ \tan \left( {\frac{{A+B}}{2}} \right)=\tan \left( {90{}^\circ -\frac{C}{2}} \right)$

$ \displaystyle \therefore \ \tan \left( {\frac{{A+B}}{2}} \right)=\cot \frac{C}{2}$

Exercise (11.3) No (2) Solution

Find the value of θ, 0° ≤ θ ≤ 360° for the following equations. Do not use table.

(a) $ \displaystyle \ \ \ \ \ \ \sin \theta =-\frac{1}{2}$

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$ \displaystyle \therefore \ \ \ \ \text{basic acute angle}=30{}^\circ $

$ \displaystyle \begin{array}{l}\ \ \ \ \ \ \text{But }\sin \theta \ \text{is negative, }\theta \ \text{may be }\\\ \ \ \ \ \ \text{either in the }{{\text{3}}^{{\text{rd}}}}\text{ or }{{\text{4}}^{{\text{th}}}}\text{ quadrant}\text{.}\end{array}$

 $ \displaystyle \therefore \,\ \ \ \theta =180{}^\circ +30{}^\circ \ (\text{or)}\ \theta =360{}^\circ -30{}^\circ \ $

$ \displaystyle \therefore \,\ \ \ \theta =210{}^\circ \ (\text{or)}\ \theta =330{}^\circ $

(b) $ \displaystyle \ \ \ \ \ \ \cos \theta =-\frac{{\sqrt{3}}}{2}$

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$ \displaystyle \therefore \ \ \ \ \text{basic acute angle}=30{}^\circ $

$ \displaystyle \begin{array}{l}\ \ \ \ \ \ \text{But }\cos \theta \ \text{is negative, }\theta \ \text{may be }\\\ \ \ \ \ \ \text{either in the }{{\text{2}}^{{\text{nd}}}}\text{ or }{{\text{3}}^{{\text{rd}}}}\text{ quadrant}\text{.}\end{array}$

$ \displaystyle \therefore \,\ \ \ \theta =180{}^\circ -30{}^\circ \ (\text{or)}\ \theta =180{}^\circ +30{}^\circ \ $

$ \displaystyle \therefore \,\ \ \ \theta =150{}^\circ \ (\text{or)}\ \theta =210{}^\circ $

(c) $ \displaystyle \ \ \ \ \ \ \cos \theta =-\frac{1}{{\sqrt{2}}}$

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$ \displaystyle \therefore \ \ \ \ \text{basic acute angle}=45{}^\circ $

$ \displaystyle \begin{array}{l}\ \ \ \ \ \ \text{But }\cos \theta \ \text{is negative, }\theta \ \text{may be }\\\ \ \ \ \ \ \text{either in the }{{\text{2}}^{{\text{nd}}}}\text{ or }{{\text{3}}^{{\text{rd}}}}\text{ quadrant}\text{.}\end{array}$

$ \displaystyle \therefore \,\ \ \ \theta =180{}^\circ -45{}^\circ \ (\text{or)}\ \theta =180{}^\circ +45{}^\circ \ $

$ \displaystyle \therefore \,\ \ \ \theta =135{}^\circ \ (\text{or)}\ \theta =225{}^\circ $

(d) $ \displaystyle \ \ \ \ \ \ \tan \theta =\sqrt{3}$

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$ \displaystyle \therefore \ \ \ \ \text{basic acute angle}=60{}^\circ $

$ \displaystyle \begin{array}{l}\ \ \ \ \ \ \text{But }\tan \theta \ \text{is positive, }\theta \ \text{may be }\\\ \ \ \ \ \ \text{either in the }{{\text{1}}^{{\text{st}}}}\text{ or }{{\text{3}}^{{\text{rd}}}}\text{ quadrant}\text{.}\end{array}$

$ \displaystyle \therefore \,\ \ \ \theta =60{}^\circ \ (\text{or)}\ \theta =180{}^\circ +60{}^\circ \ $

$ \displaystyle \therefore \,\ \ \ \theta =60{}^\circ \ (\text{or)}\ \theta =240{}^\circ $

(e) $ \displaystyle \ \ \ \ \ \ \tan 2\theta =1$

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$ \displaystyle \therefore \ \ \ \ \text{basic acute angle}=45{}^\circ $

$ \displaystyle \begin{array}{l}\ \ \ \ \ \ \text{But }\tan 2\theta \ \text{is positive, }2\theta \ \text{may be }\\\ \ \ \ \ \ \text{either in the }{{\text{1}}^{{\text{st}}}}\text{ or }{{\text{3}}^{{\text{rd}}}}\text{ quadrant}\text{.}\end{array}$

$ \displaystyle \underline{{\text{For }{{\text{1}}^{{\text{st}}}}\ \text{quadrant}}}\text{,}$

$ \displaystyle \therefore \,\ \ \ 2\theta =45{}^\circ \ (\text{or)}\ 2\theta =360{}^\circ +45{}^\circ \ $

$ \displaystyle \ \ \ \ \ 2\theta =45{}^\circ \ (\text{or)}\ 2\theta =405{}^\circ \ $

$ \displaystyle \ \ \ \ \ \theta =22{}^\circ 3{0}'(\text{or)}\ \theta =\ 202{}^\circ 3{0}'$

$ \displaystyle \underline{{\text{For }{{\text{3}}^{{\text{rd}}}}\ \text{quadrant}}}\text{,}$

$ \displaystyle \ \,\ \ \ 2\theta =180{}^\circ +45{}^\circ \ (\text{or)}\ 2\theta =360{}^\circ +180{}^\circ +45{}^\circ $

$ \displaystyle \ \ \ \ \ 2\theta =225{}^\circ \ (\text{or)}\ 2\theta =585{}^\circ $

$ \displaystyle \ \ \ \ \ \theta =112{}^\circ 3{0}'(\text{or)}\ \theta =292{}^\circ 3{0}'$

(f) $ \displaystyle \ \ \ \ \ \ \tan 3\theta =-1$

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$ \displaystyle \therefore \ \ \ \ \text{basic acute angle}=45{}^\circ $

$ \displaystyle \begin{array}{l}\ \ \ \ \ \ \text{But }\tan 3\theta \ \text{is negative, 3}\theta \ \text{may be }\\\ \ \ \ \ \ \text{either in the }{{\text{2}}^{{\text{nd}}}}\text{ or }{{\text{4}}^{{\text{th}}}}\text{ quadrant}\text{.}\end{array}$

$ \displaystyle \underline{{\text{For }{{\text{2}}^{{\text{nd}}}}\ \text{quadrant}}}\text{,}$

$ \displaystyle \ \,\ \ \ 3\theta =180{}^\circ -45{}^\circ \ $

$ \displaystyle \ \,\ \ \ 3\theta =135{}^\circ \ $

$ \displaystyle \ \,\ \ \ \theta =45{}^\circ \ $

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ (\text{or)}$

$ \displaystyle \ \ \ \ \ 3\theta =360{}^\circ +180{}^\circ -45{}^\circ \ $

$ \displaystyle \ \,\ \ \ 3\theta =495{}^\circ \ $

$ \displaystyle \ \,\ \ \ \theta =165{}^\circ \ $

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ (\text{or)}\ $

$ \displaystyle \ \ \ \ \ 3\theta =360{}^\circ +360{}^\circ +180{}^\circ -45{}^\circ \ $

$ \displaystyle \ \,\ \ \ 3\theta =855{}^\circ \ $

$ \displaystyle \ \,\ \ \ \theta =285{}^\circ $

$ \displaystyle \underline{{\text{For }{{\text{4}}^{{\text{th}}}}\ \text{quadrant}}}\text{,}$

$ \displaystyle \ \,\ \ \ 3\theta =360{}^\circ -45{}^\circ $

$ \displaystyle \ \,\ \ \ 3\theta =315{}^\circ \ $

$ \displaystyle \ \,\ \ \ \theta =105{}^\circ \ $

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ (\text{or)}\ $

$ \displaystyle \ \ \ \ \ 3\theta =360{}^\circ +360{}^\circ -45{}^\circ $

$ \displaystyle \ \,\ \ \ 3\theta =675{}^\circ \ $

$ \displaystyle \ \,\ \ \ \theta =225{}^\circ \ \ $

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ (\text{or)}\ $

$ \displaystyle \ \ \ \ \ 3\theta =360{}^\circ +360{}^\circ +360{}^\circ -45{}^\circ \ $

$ \displaystyle \ \,\ \ \ 3\theta =1035{}^\circ \ $

$ \displaystyle \ \,\ \ \ \theta =345{}^\circ \ \ $

diagram မ်ားသည္ principal angle မ်ား ကို ဆံုးျဖတ္ရာတြင္  အေထာက္အကူ ျဖစ္ေစရန္ ေရးဆြဲ ေဖၚျပျခင္း ျဖစ္သည္။ နားလည္ ကၽြမ္းက်င္ သြားလ်င္ diagram မ်ား မေရးဆြဲပဲ ေျဖဆိုႏိုင္ ပါသည္။ က်န္ေသာပုစာၦမ်ားအတြက္ diagram မဆြဲေတာ့ပဲ ေျဖဆိုပါမည္။

(g) $ \displaystyle \ \ \ \ \ \ \tan (3\theta -30{}^\circ )=-1$

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$ \displaystyle \begin{array}{l}\therefore \ \ \ \ \text{basic acute angle}=45{}^\circ \\\\\ \ \ \ \ \ \text{But }\tan (3\theta -30{}^\circ )\ \text{is negative, }(3\theta -30{}^\circ )\ \text{may be }\\\ \ \ \ \ \ \text{either in the }{{\text{2}}^{{\text{nd}}}}\text{ or }{{\text{4}}^{{\text{th}}}}\text{ quadrant}\text{.}\\\\\underline{{\text{For }{{\text{2}}^{{\text{nd}}}}\ \text{quadrant}}}\text{,}\\\\\ \,\ \ \ 3\theta -30{}^\circ =180{}^\circ -45{}^\circ \\\ \\\ \,\ \ \ 3\theta =165{}^\circ \ \\\\\ \,\ \ \ \theta =55{}^\circ \ \\\\\ \ \ \ \ \ \ \ \ \ \ (\text{or)}\ \\\\\ \ \ \ \ 3\theta -30{}^\circ =360{}^\circ +180{}^\circ -45{}^\circ \ \\\\\ \,\ \ \ 3\theta =525{}^\circ \ \\\\\ \,\ \ \ \theta =175{}^\circ \ \\\\\ \ \ \ \ \ \ \ \ \ \ (\text{or)}\ \\\\\ \ \ \ \ 3\theta -30{}^\circ =360{}^\circ +360{}^\circ +180{}^\circ -45{}^\circ \ \\\\\ \,\ \ \ 3\theta =885{}^\circ \ \\\\\ \,\ \ \ \theta =295{}^\circ \ \ \ \\\\\underline{{\text{For }{{\text{4}}^{{\text{th}}}}\ \text{quadrant}}}\text{,}\\\\\ \,\ \ \ 3\theta -30{}^\circ =360{}^\circ -45{}^\circ \\\\\ \,\ \ \ 3\theta =345{}^\circ \ \\\\\ \,\ \ \ \theta =115{}^\circ \ \ \\\\\ \ \ \ \ \ \ \ \ \ \ (\text{or)}\ \\\\\ \ \ \ \ 3\theta -30{}^\circ =360{}^\circ +360{}^\circ -45{}^\circ \\\\\ \,\ \ \ 3\theta =705{}^\circ \ \\\\\ \,\ \ \ \theta =295{}^\circ \ \ \\\\\ \ \ \ \ \ \ \ \ \ \ (\text{or)}\ \\\\\ \ \ \ \ 3\theta -30{}^\circ =360{}^\circ +360{}^\circ +360{}^\circ -45{}^\circ \ \\\\\ \,\ \ \ 3\theta =1065{}^\circ \ \\\\\ \,\ \ \ \theta =355{}^\circ \ \ \end{array}$

(h) $ \displaystyle \ \ \ \ \ \ \cos 2\theta =-\frac{1}{2}$

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$ \displaystyle \begin{array}{l}\therefore \ \ \ \ \text{basic acute angle}=60{}^\circ \\\\\ \ \ \ \ \ \text{But }\cos 2\theta \ \text{is negative, }2\theta \ \text{may be }\\\ \ \ \ \ \ \text{either in the }{{\text{2}}^{{\text{nd}}}}\text{ or }{{\text{3}}^{{\text{rd}}}}\text{ quadrant}\text{.}\\\\\underline{{\text{For }{{\text{2}}^{{\text{nd}}}}\ \text{quadrant}}}\text{,}\\\\\ \,\ \ \ 2\theta =180{}^\circ -60{}^\circ \\\ \\\ \,\ \ \ 2\theta =120{}^\circ \ \\\\\ \,\ \ \ \theta =60{}^\circ \ \\\\\ \ \ \ \ \ \ \ \ \ \ (\text{or)}\ \\\\\ \ \ \ \ 2\theta =360{}^\circ +180{}^\circ -60{}^\circ \ \\\\\ \,\ \ \ 2\theta =480{}^\circ \ \\\\\ \,\ \ \ \theta =240{}^\circ \ \\\\\underline{{\text{For }{{\text{3}}^{{\text{rd}}}}\ \text{quadrant}}}\text{,}\\\\\ \,\ \ \ 2\theta =180{}^\circ +60{}^\circ \\\\\ \,\ \ \ 2\theta =240{}^\circ \ \\\\\ \,\ \ \ \theta =120{}^\circ \ \ \\\\\ \ \ \ \ \ \ \ \ \ \ (\text{or)}\ \\\\\ \ \ \ \ 2\theta =360{}^\circ +180{}^\circ +60{}^\circ \\\\\ \,\ \ \ 2\theta =600{}^\circ \ \\\\\ \,\ \ \ \theta =300{}^\circ \ \ \ \ \end{array}$

(i) $ \displaystyle \ \ \ \ \ \sin (2\theta +30{}^\circ )=\frac{{\sqrt{3}}}{2}$

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$ \displaystyle \begin{array}{l}\therefore \ \ \ \ \text{basic acute angle}=60{}^\circ \\\\\ \ \ \ \ \ \text{But }\sin (2\theta +30{}^\circ )\ \text{is positive, }2\theta +30{}^\circ \ \text{may be }\\\ \ \ \ \ \ \text{either in the }{{\text{1}}^{{\text{st}}}}\text{ or }{{\text{2}}^{{\text{nd}}}}\text{ quadrant}\text{.}\\\\\underline{{\text{For }{{\text{1}}^{{\text{st}}}}\ \text{quadrant}}}\text{,}\\\\\ \,\ \ \ 2\theta +30{}^\circ =60{}^\circ \\\ \\\ \,\ \ \ 2\theta =30{}^\circ \ \\\\\ \,\ \ \ \theta =15{}^\circ \ \\\\\ \ \ \ \ \ \ \ \ \ \ (\text{or)}\ \\\\\ \ \ \ \ 2\theta =360{}^\circ +60{}^\circ \ \\\\\ \,\ \ \ 2\theta =420{}^\circ \ \\\\\ \,\ \ \ \theta =210{}^\circ \ \\\\\underline{{\text{For }{{\text{2}}^{{\text{nd}}}}\ \text{quadrant}}}\text{,}\\\\\ \,\ \ \ 2\theta +30{}^\circ =180{}^\circ -60{}^\circ \\\ \\\ \,\ \ \ 2\theta =120{}^\circ \ \\\\\ \,\ \ \ \theta =60{}^\circ \ \\\\\ \ \ \ \ \ \ \ \ \ \ (\text{or)}\ \\\\\ \ \ \ \ 2\theta =360{}^\circ +180{}^\circ -60{}^\circ \ \\\\\ \,\ \ \ 2\theta =480{}^\circ \ \\\\\ \,\ \ \ \theta =240{}^\circ \ \end{array}$

(j) $ \displaystyle \ \ \ \ \ \ \tan \frac{1}{2}\theta =-\frac{1}{{\sqrt{3}}}$

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$ \displaystyle \therefore \ \ \ \ \text{basic acute angle}=30{}^\circ $

$ \displaystyle \ \ \ \ \ \ \text{But}\ \tan \frac{1}{2}\theta \ \text{is positive, }\frac{1}{2}\theta \ \text{may be }$

$ \displaystyle \ \ \ \ \ \ \text{either in the }{{\text{2}}^{{\text{nd}}}}\text{ or }{{\text{4}}^{{\text{th}}}}\text{ quadrant}\text{.}$

$ \displaystyle \therefore \ \ \ \ \frac{1}{2}\theta =180{}^\circ -30{}^\circ \ (\text{or})\ \frac{1}{2}\theta =360{}^\circ -30{}^\circ $

$ \displaystyle \therefore \ \ \ \ \frac{1}{2}\theta =150{}^\circ \ (\text{or})\ \frac{1}{2}\theta =330{}^\circ $

$ \displaystyle \therefore \ \ \ \ \theta =300{}^\circ \ (\text{or})\ \theta =660{}^\circ $

$ \displaystyle \begin{array}{l}\ \ \ \ \ \ \text{But}\ 0{}^\circ \le \theta \le 360{}^\circ ,\theta =660{}^\circ \ \text{is not in domain}\text{.}\\\\\therefore \ \ \ \ \theta =300{}^\circ \ \text{is the only solution}\text{.}\end{array}$

(k) $ \displaystyle \ \ \ \ \ \ \sin \theta =0.6521$

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$ \displaystyle \begin{array}{l}\therefore \ \ \ \ \text{basic acute angle}=40{}^\circ 4{2}'\ (\text{using table)}\\\\\ \ \ \ \ \ \text{But}\ \sin \theta \ \text{is positive, }\theta \ \text{may be }\\\ \ \ \ \ \ \text{either in the }{{\text{1}}^{{\text{st}}}}\text{ or }{{\text{2}}^{{\text{nd}}}}\text{ quadrant}\text{.}\\\\\therefore \ \ \ \ \theta =40{}^\circ 4{2}'\ (\text{or})\ \theta =180{}^\circ -40{}^\circ 4{2}'\\\\\therefore \ \ \ \ \theta =40{}^\circ 4{2}'\ (\text{or})\ \theta =139{}^\circ 1{8}'\\\ \end{array}$

(l) $ \displaystyle \ \ \ \ \ \ \cos \theta =-0.3854$

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$ \displaystyle \begin{array}{l}\therefore \ \ \ \ \text{basic acute angle}=67{}^\circ 2{0}'\ (\text{using table)}\\\\\ \ \ \ \ \ \text{But}\ \cos \theta \ \text{is negative, }\theta \ \text{may be }\\\ \ \ \ \ \ \text{either in the }{{\text{2}}^{{\text{nd}}}}\text{ or }{{\text{3}}^{{\text{rd}}}}\text{ quadrant}\text{.}\\\\\therefore \ \ \ \ \theta =180{}^\circ -67{}^\circ 2{0}'\ (\text{or})\ \theta =180{}^\circ +67{}^\circ 2{0}'\\\\\therefore \ \ \ \ \theta =112{}^\circ 4{0}'\ (\text{or})\ \theta =247{}^\circ 2{0}'\end{array}$

Exercise (11.3) No (3) Solution

Solve the following equations for 0° ≤ x ≤ 360°.

$ \displaystyle {(\text{a})\ 2\sin x\cos x=\sin x}$

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$ \displaystyle {\ \ \ \ \ 2\sin x\cos x-\sin x=0}$

$ \displaystyle {\ \ \ \ \ \sin x(2\cos x-1)=0}$

$ \displaystyle \ \ \ \ \ \sin x=0\ \ (\text{or})\ \cos x=\frac{1}{2}$

$ \displaystyle \begin{array}{l}(\text{i})\ \ \text{For }\sin x=0,\\\\\ \ \ \ \ \ x=0{}^\circ \ (\text{or})\ x=180{}^\circ \ (\text{or})\ x=360{}^\circ \end{array}$

$ \displaystyle (\text{ii})\ \text{For }\cos x=\frac{1}{2},$

$ \displaystyle \begin{array}{l}\ \ \ \ \ \ x=60{}^\circ \ (\text{or})\ x=360{}^\circ -60{}^\circ \\\\\ \ \ \ \ \ x=60{}^\circ \ (\text{or})\ x=300{}^\circ \end{array}$

$ \displaystyle (\text{b})\ 3\tan x\sin x=2\tan x$

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$ \displaystyle \begin{array}{l}\ \ \ \ \ 3\tan x\sin x-2\tan x=0\\\\\ \ \ \ \ \tan x(3\sin x-2)=0\\\\\ \ \ \ \ \tan x=0\ \ (\text{or})\ \sin x=\displaystyle \frac{2}{3}\\\\(\text{i})\ \ \text{For }\tan x=0,\\\\\ \ \ \ \ \ x=0{}^\circ \ (\text{or})\ x=180{}^\circ \ (\text{or})\ x=360{}^\circ \\\\(\text{ii})\ \text{For }\sin x=\displaystyle \frac{2}{3}=0.6667,\\\\\ \ \ \ \ \ x=41{}^\circ 4{9}'\ (\text{or})\ x=180{}^\circ -41{}^\circ 4{9}'{}^\circ \\\\\ \ \ \ \ \ x=41{}^\circ 4{9}'\ (\text{or})\ x=138{}^\circ 1{1}'\end{array}$

$ \displaystyle (\text{c})\ 3\ {{\sin }^{2}}x=4\sin x$

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$ \displaystyle \begin{array}{l}\ \ \ \ \ 3{{\sin }^{2}}x-4\sin x=0\\\\\ \ \ \ \ \sin x(3\sin x-4)=0\\\\\ \ \ \ \ \sin x=0\ \ (\text{or})\ \sin x=\displaystyle \frac{4}{3}\\\\(\text{i})\ \ \text{For }\sin x=0,\\\\\ \ \ \ \ \ x=0{}^\circ \ (\text{or})\ x=180{}^\circ \ (\text{or})\ x=360{}^\circ \\\\(\text{ii})\ \text{For }\sin x=\displaystyle \frac{4}{3}=1.333\\\ \\\ \ \ \ \ \text{Since }-1\le \sin x\le 1,\\\\\ \ \ \ \text{sin }x=\displaystyle \frac{4}{3}\text{ is impossible}\text{.}\end{array}$

$ \displaystyle (\text{d})\ 5\sin x\cos x=2\cos x$

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$ \displaystyle \begin{array}{l}\ \ \ \ \ 5\sin x\cos x-2\cos x=0\\\\\ \ \ \ \ \cos x(5\sin x-2)=0\\\\\ \ \ \ \ \cos x=0\ \ (\text{or})\ \sin x=\displaystyle \frac{2}{5}\\\\(\text{i})\ \ \text{For }\cos x=0,\\\\\ \ \ \ \ \ x=90{}^\circ \ (\text{or})\ x=270{}^\circ \ \\\\(\text{ii})\ \text{For }\sin x=\displaystyle \frac{2}{5}=0.4\\\ \\\ \ \ \ x=23{}^\circ 3{5}'\ (\text{or})\ x=180{}^\circ -23{}^\circ 3{5}'\end{array}$

$ \displaystyle (\text{e})\ {{\cos }^{2}}x-\cos x=2$

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$ \displaystyle \begin{array}{l}\ \ \ \ \ {{\cos }^{2}}x-\cos x-2=0\\\\\ \ \ \ \ (\cos x+1)(\cos x-2)=0\\\\\ \ \ \ \ \cos x=-1\ \ (\text{or})\ \cos x=2\\\\(\text{i})\ \ \text{For }\cos x=-1,\\\\\ \ \ \ \ \ x=180{}^\circ \ \\\\(\text{ii})\ \text{For }\cos x=2\\\ \\\ \ \ \ \ \text{Since }-1\le \cos x\le 1,\\\ \ \\\ \ \ \ \ \cos x=2\ \text{is not in domain}\text{.}\\\\\therefore \ \ \ x=180{}^\circ \ \text{is the only solution}\text{.}\end{array}$

$ \displaystyle (\text{f})\ 2\sin x\cos x-\cos x=0$

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$ \displaystyle \begin{array}{l}\ \ \ \ \ \cos x(2\sin x-1)=0\\\\\ \ \ \ \ \cos x=0\ \ (\text{or})\ \sin x=\displaystyle \frac{1}{2}\\\\(\text{i})\ \ \text{For }\cos x=0,\\\\\ \ \ \ \ \ x=90{}^\circ \ (\text{or})\ x=270{}^\circ \\\\(\text{ii})\ \text{For }\sin x=\displaystyle \frac{1}{2}\\\ \\\ \ \ \ \ x=30{}^\circ \ (\text{or})\ x=180{}^\circ -30{}^\circ \\\\\ \ \ \ \ x=30{}^\circ \ (\text{or})\ x=150{}^\circ \end{array}$

$ \displaystyle (\text{g})\ 2{{\sin }^{2}}x-\sin x=1$

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$ \displaystyle \begin{array}{l}\ \ \ \ \ 2{{\sin }^{2}}x-\sin x-1=0\\\\\ \ \ \ \ (2\sin x+1)(\sin x-1)=0\\\\\ \ \ \ \ \sin x=\displaystyle - \frac{1}{2}\ \ (\text{or})\ \sin x=1\\\\(\text{i})\ \ \text{For }\ \sin x=\displaystyle -\frac{1}{2},\\\\\ \ \ \ \ \ x=180{}^\circ +30{}^\circ \ (\text{or})\ x=360{}^\circ -30{}^\circ \\\\\ \ \ \ \ \ x=210{}^\circ \ (\text{or})\ x=330{}^\circ \\\\(\text{ii})\ \text{For }\sin x=1\\\ \\\ \ \ \ \ x=90{}^\circ \ \end{array}$

$ \displaystyle (\text{h})\ 2\sin x\cos x=\sqrt{3}\cos x$

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$ \displaystyle \begin{array}{l}\ \ \ \ \ 2\sin x\cos x-\sqrt{3}\cos x=0\\\\\ \ \ \ \ \cos x(2\sin x-\sqrt{3})=0\\\\\ \ \ \ \ \cos x=0\ \ (\text{or})\ \sin x=\displaystyle \frac{{\sqrt{3}}}{2}\\\\(\text{i})\ \ \text{For }\ \cos x=0,\\\\\ \ \ \ \ \ x=90{}^\circ \ (\text{or})\ x=270{}^\circ \\\\(\text{ii})\ \text{For }\sin x=\displaystyle \frac{{\sqrt{3}}}{2}\\\ \\\ \ \ \ \ x=60{}^\circ \ (\text{or})\ x=180{}^\circ -60{}^\circ \\\\\ \ \ \ \ x=60{}^\circ \ (\text{or})\ x=120{}^\circ \end{array}$

$ \displaystyle (\text{i})\ 2\sin x\cos x-\cos x+4\sin x-2=0$

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$ \displaystyle \begin{array}{l}\ \ \ \ \ \cos x(2\sin x-1)+2(2\sin x-1)=0\\\\\ \ \ \ \ (2\sin x-1)(\cos x+2)=0\\\\\ \ \ \ \ \sin x=\displaystyle \frac{1}{2}\ \ (\text{or})\ \cos x=-2\\\\(\text{i})\ \ \text{For }\ \sin x=\displaystyle \frac{1}{2},\\\\\ \ \ \ \ \ x=30{}^\circ \ (\text{or})\ x=180{}^\circ -30{}^\circ \\\\\ \ \ \ \ \ x=30{}^\circ \ (\text{or})\ x=150{}^\circ \\\\(\text{ii})\ \text{For }\cos x=-2,\\\\\ \ \ \ \ \text{Since }-1\le \cos x\le 1,\ \\\\\ \ \ \ \ \cos x=-2\,\text{is}\ \text{out of domain and}\ \\\ \ \ \ \ \text{hence }\cos x=-2\ \text{has no solution}\text{.}\end{array}$

$ \displaystyle (\text{j})\ \ 8{{\cos }^{2}}x-2\cos x-5=\sec x$

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$ \displaystyle \begin{array}{l}\ \ \ \ \ 8{{\cos }^{2}}x-2\cos x-5=\displaystyle \frac{1}{{\cos x}}\\\\\ \ \ \ \ 8{{\cos }^{3}}x-2{{\cos }^{2}}x-5\cos x=1\\\\\ \ \ \ \ 8{{\cos }^{3}}x-2{{\cos }^{2}}x-5\cos x-1=0\\\\\ \ \ \ \ (2\cos x+1)(4\cos x+1)(x-1)=0\ \left[ {\text{using factor theorem}} \right]\\\\\therefore \ \ \ \cos x=\displaystyle -\frac{1}{2}\ (\text{or})\ \cos x=\displaystyle -\frac{1}{4}\ (\text{or})\ \cos x=1\\\\(\text{i})\ \ \text{For }\ \cos x=\displaystyle -\frac{1}{2},\\\\\ \ \ \ \ \ x=180{}^\circ -60{}^\circ (\text{or})\ x=180{}^\circ +60{}^\circ \\\\\ \ \ \ \ \ x=120{}^\circ \ (\text{or})\ x=240{}^\circ \\\\(\text{ii})\ \text{For }\cos x=\displaystyle -\frac{1}{4}=0.25,\\\\\ \ \ \ \ \ x=180{}^\circ -14{}^\circ 2{9}'\ (\text{or})\ x=180{}^\circ +14{}^\circ 2{9}'\\\\\ \ \ \ \ \ x=165{}^\circ 3{1}'\ (\text{or})\ x=194{}^\circ 2{9}'\\\\(\text{iii})\ \text{For }\ \cos x=0,\\\\\ \ \ \ \ \ x=90{}^\circ (\text{or})\ x=270{}^\circ \end{array}$

Exercise (11.3) - No(1) Sample Solutions

(b) Solution (1)

$ \displaystyle \sin \text{135}{}^\circ =\sin (\text{180}{}^\circ -45{}^\circ )=\sin 45{}^\circ =\frac{{\sqrt{2}}}{2}$

$ \displaystyle \cos 135{}^\circ \text{ =}\cos \text{(180}{}^\circ -45{}^\circ )=-\cos 45{}^\circ =-\frac{{\sqrt{2}}}{2}$

$ \displaystyle \begin{array}{l}\tan 135{}^\circ \text{ =}-1\\\\\cot 135{}^\circ \text{ =}-1\\\\\sec 135{}^\circ \text{ =}-\sqrt{2}\\\\\operatorname{cosec}135{}^\circ \text{ =}\sqrt{2}\end{array}$

(b) Solution (2)

$ \displaystyle \begin{array}{l}\ \ \ \text{principal angle = 135}{}^\circ \\\\\therefore \text{basic acute angle = 45}{}^\circ \end{array}$

$ \displaystyle \therefore \ \sin \text{45}{}^\circ =\frac{{\sqrt{2}}}{2}\operatorname{and}\ \cos \text{45}{}^\circ =\frac{{\sqrt{2}}}{2}$

$ \displaystyle \begin{array}{l}\ \ \ \sin \text{135}{}^\circ =\sin \text{45}{}^\circ \operatorname{and}\ \cos \text{135}{}^\circ =\cos \text{45}{}^\circ \ \text{numerically}\text{.}\\\\\ \ \ \text{But 135}{}^\circ \ \text{lies in the second quadrant}\text{.}\end{array}$

$ \displaystyle \therefore \sin \text{135}{}^\circ =\sin \text{45}{}^\circ =\frac{{\sqrt{2}}}{2}$

$ \displaystyle \ \ \ \cos \text{135}{}^\circ =-\cos \text{45}{}^\circ =-\frac{{\sqrt{2}}}{2}$

$ \displaystyle \ \ \ \begin{array}{*{20}{l}} {\tan {{{135}}^{{}^\circ }}\text{ =}-1} \\ {} \\ {\cot {{{135}}^{{}^\circ }}\text{ =}-1} \\ {} \\ {\sec {{{135}}^{{}^\circ }}\text{ =}-\sqrt{2}} \\ {} \\ {\operatorname{cosec}{{{135}}^{{}^\circ }}\text{ =}\sqrt{2}} \end{array}$

(d) Solution (1)

$ \displaystyle \ \ \ \sin 210{}^\circ =\sin (180{}^\circ +30{}^\circ )=-\sin 30{}^\circ =\displaystyle -\frac{1}{2}$

$ \displaystyle \ \ \ \cos 210{}^\circ =\cos (180{}^\circ +30{}^\circ )=-\cos 30{}^\circ =-\frac{{\sqrt{3}}}{2}$

$ \displaystyle \ \ \ \tan 210{}^\circ =\displaystyle \frac{{\sqrt{3}}}{3}$

$ \displaystyle \ \ \ \cot 210{}^\circ =\sqrt{3}$

$ \displaystyle \ \ \ \sec 210{}^\circ =\displaystyle -\frac{{2\sqrt{3}}}{3}$

$ \displaystyle \ \ \ \operatorname{cosec} 210{}^\circ =-2$

(d) Solution (2)

$ \displaystyle \ \ \ \text{principal angle = 210}{}^\circ $

$ \displaystyle \therefore \text{basic acute angle = 30}{}^\circ $

$ \displaystyle \therefore \ \sin \text{30}{}^\circ =\frac{1}{2}\operatorname{and}\ \cos \text{30}{}^\circ =\frac{{\sqrt{3}}}{2}$

$ \displaystyle \ \ \ \sin \text{210}{}^\circ =\sin \text{30}{}^\circ \operatorname{and}\ \cos \text{210}{}^\circ =\cos \text{30}{}^\circ \ \text{numerically}\text{.}$

$ \displaystyle \ \ \ \text{But 210}{}^\circ \ \text{lies in the third quadrant}\text{.}$

$ \displaystyle \therefore \sin \text{210}{}^\circ =\sin \text{30}{}^\circ =-\frac{1}{2}$

$ \displaystyle \ \ \ \cos \text{210}{}^\circ =-\cos \text{30}{}^\circ =\displaystyle -\frac{{\sqrt{3}}}{2}$

$ \displaystyle \ \ \ \begin{array}{*{20}{l}} {\tan \text{210}{}^\circ \text{ =}\displaystyle \frac{{\sqrt{3}}}{3}} \\ {} \\ {\cot \text{210}{}^\circ \text{ =}\sqrt{3}} \\ {} \\ {\sec \text{210}{}^\circ \text{ =}\displaystyle -\frac{{2\sqrt{3}}}{3}} \\ {} \\ {\operatorname{cosec}\text{210}{}^\circ \text{ =}-2} \end{array}$

(m) Solution (1)

$ \displaystyle \ \ \ \sin (-120{}^\circ )=-\sin 120{}^\circ $

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =-\sin (180{}^\circ -60{}^\circ )$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =-\sin 60{}^\circ $

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =-\frac{{\sqrt{3}}}{2}$

$ \displaystyle \ \ \ \ \cos (-120{}^\circ )=\cos 120{}^\circ $

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\cos (180{}^\circ -60{}^\circ )$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =-\cos 60{}^\circ $

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =-\frac{1}{2}$

$ \displaystyle \ \ \ \tan (-120{}^\circ )=\sqrt{3}$

$ \displaystyle \ \ \ \cot (-120{}^\circ )=\frac{{\sqrt{3}}}{3}$

$ \displaystyle \ \ \ \sec (-120{}^\circ )=-2$

$ \displaystyle \ \ \ \operatorname{cosec}(-120{}^\circ )=-\frac{{2\sqrt{3}}}{3}$

(m) Solution (2)

$ \displaystyle \ \ \ \text{principal angle = }-120{}^\circ $

$ \displaystyle \therefore \text{basic acute angle = 60}{}^\circ $

$ \displaystyle \therefore \ \sin \text{60}{}^\circ =\frac{{\sqrt{3}}}{2}\operatorname{and}\ \cos \text{60}{}^\circ =\frac{1}{2}$

$ \displaystyle \ \ \ \sin (-120{}^\circ )=\sin \text{60}{}^\circ \operatorname{and}\ \cos (-120{}^\circ)x =\cos \text{60}{}^\circ \ \text{numerically}\text{.}$

$ \displaystyle \ \ \ \text{But }(-120{}^\circ )\ \text{lies in the third quadrant}\text{.}$

$ \displaystyle \therefore \sin (-120{}^\circ )=\sin \text{60}{}^\circ =-\frac{{\sqrt{3}}}{2}$

$ \displaystyle \ \ \ \cos (-120{}^\circ )=-\cos \text{60}{}^\circ =-\frac{1}{2}$

$ \displaystyle \ \ \ \tan (-120{}^\circ )=\sqrt{3}$

$ \displaystyle \ \ \ \cot (-120{}^\circ )=\frac{{\sqrt{3}}}{3}$

$ \displaystyle \ \ \ \sec (-120{}^\circ )=-2$

$ \displaystyle \ \ \ \operatorname{cosec}(-120{}^\circ )=-\frac{{2\sqrt{3}}}{3}$

(q) Solution (1)

$ \displaystyle \ \ \ \sin 480{}^\circ =\sin (360{}^\circ +60{}^\circ )$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\sin 60{}^\circ $

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{\sqrt{3}}}{2}$

$ \displaystyle \ \ \ \ \cos 480{}^\circ =\cos (360{}^\circ +60{}^\circ )$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\cos 60{}^\circ $

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{1}{2}$

$ \displaystyle \ \ \ \tan 480{}^\circ =\sqrt{3}$

$ \displaystyle \ \ \ \cot 480{}^\circ =\frac{{\sqrt{3}}}{3}$

$ \displaystyle \ \ \ \sec 480{}^\circ =2$

$ \displaystyle \ \ \ \operatorname{cosec}480{}^\circ =\frac{{2\sqrt{3}}}{3}$

(q) Solution (2)

$ \displaystyle \ \ \ \text{principal angle = }480{}^\circ $

$ \displaystyle \therefore \text{basic acute angle = 60}{}^\circ $

$ \displaystyle \therefore \ \sin \text{60}{}^\circ =\frac{{\sqrt{3}}}{2}\operatorname{and}\ \cos \text{60}{}^\circ =\frac{1}{2}$

$ \displaystyle \ \ \ \sin 480{}^\circ =\sin \text{60}{}^\circ \operatorname{and}\ \cos 480{}^\circ =\cos \text{60}{}^\circ \ \text{numerically}\text{.}$

$ \displaystyle \ \ \ \text{But}\ 480{}^\circ \ \text{lies in the first quadrant}\text{.}$

$ \displaystyle \therefore \sin 480{}^\circ =\sin \text{60}{}^\circ =\frac{{\sqrt{3}}}{2}$

$ \displaystyle \ \ \ \cos 480{}^\circ =\cos \text{30}{}^\circ =\frac{1}{2}$

$ \displaystyle \ \ \ \tan 480{}^\circ =\sqrt{3}$

$ \displaystyle \ \ \ \cot 480{}^\circ =\frac{{\sqrt{3}}}{3}$

$ \displaystyle \ \ \ \sec 480{}^\circ =2$

$ \displaystyle \ \ \ \operatorname{cosec}480{}^\circ =\frac{{2\sqrt{3}}}{3}$

Trigonometric Ratios of 0°, 180°, 270° and 360°

Trigonometric Ratios of 0° 

From the unit circle we have ,

$ \displaystyle \sin 0°=y=0$

$ \displaystyle \cos 0°=x=1$

Therefore

$ \displaystyle \tan 0°=\frac{y}{x}=\frac{0}{1}=1$

$ \displaystyle \cot 0°=\frac{x}{y}=\frac{1}{0}=\text{undefined}$

$ \displaystyle \sec 0°=\frac{1}{x}=\frac{1}{1}=1$

$ \displaystyle \operatorname{cosec} 0°=\frac{1}{y}=\frac{1}{0}=\text{undefined}$

Trigonometric Ratios of 90° 

Similarly,

$ \displaystyle \sin 90°=y=1$

$ \displaystyle \cos 90°=x=0$

$ \displaystyle \tan 90°=\frac{y}{x}=\frac{1}{0}=\text{undefined}$

$ \displaystyle \cot 90°=\frac{x}{y}=\frac{0}{1}=0$

$ \displaystyle \sec 90°=\frac{1}{x}=\frac{1}{0}=\text{undefined}$

$ \displaystyle \operatorname{cosec} 90°=\frac{1}{y}=\frac{1}{1}=1$

Trigonometric Ratios of 180° 

Similarly,

$ \displaystyle \sin 180°=y=0$

$ \displaystyle \cos 180°=x=-1$

$ \displaystyle \tan 180°=\frac{y}{x}=\frac{0}{-1}=0$

$ \displaystyle \cot 180°=\frac{x}{y}=\frac{-1}{0}=\text{undefined}$

$ \displaystyle \sec 180°=\frac{1}{x}=\frac{1}{-1}=-1$

$ \displaystyle \operatorname{cosec} 180°=\frac{1}{y}=\frac{1}{0}=\text{undefined}$

Trigonometric Ratios of 270° 

Similarly,

$ \displaystyle \sin 270°=y=-1$

$ \displaystyle \cos 270°=x=0$

$ \displaystyle \tan 270°=\frac{y}{x}=\frac{-1}{0}=\text{undefined}$

$ \displaystyle \cot 270°=\frac{x}{y}=\frac{0}{-1}=0$

$ \displaystyle \sec 270°=\frac{1}{x}=\frac{1}{0}=\text{undefined}$

$ \displaystyle \operatorname{cosec} 270°=\frac{1}{y}=\frac{1}{-1}=-1$

Trigonometric Ratios of 360° 

Similarly,

$ \displaystyle \sin 360°=y=0$

$ \displaystyle \cos 360°=x=1$

$ \displaystyle \tan 360°=\frac{y}{x}=\frac{0}{1}=1$

$ \displaystyle \cot 360°=\frac{x}{y}=\frac{1}{0}=\text{undefined}$

$ \displaystyle \sec 360°=\frac{1}{x}=\frac{1}{1}=1$

$ \displaystyle \operatorname{cosec} 360°=\frac{1}{y}=\frac{1}{0}=\text{undefined}$


Applet တြင္ undefined အတြက္ ∞ သေကၤတ ကို သံုးထားပါသည္။

Trigonometric Ratios of Special Angles

အနား တစ္ဖက္ $ \displaystyle x$ unit ရွိတဲ့ စတုရနး္ $ \displaystyle ABCD$ ဆိုပါစို႔။

စတုရန္း ျဖစ္ေသာေၾကာင့္ ေထာင့္ျဖတ္မ်ဥ္း $ \displaystyle AC$ က သက္ဆိုင္ရာ ေထာင့္မ်ားကို ထက္၀က္ပိုင္းပါတယ္။ ဒါဆိုရင္ရင္ ပံုမွာ ျမင္ေတြ႔ရတဲ့ အတိုင္း ထပ္တူညီ ေထာင့္မွန္ ႀတိဂံႏွစ္ခု ျဖစ္လာပါတယ္။

၎တို႔အထဲက ေထာင့္မွန္ႀတိဂံ $ \displaystyle ABC$ ကို ခြဲထုတ္လိုက္ရင္ $ \displaystyle \vartriangle ABC$ ဟာ ႏွစ္နားညီ တဲ့ ေတာင့္မွန္ ႀတိဂံတစ္ခု  ($ \displaystyle 45°-45°$ right triangle လို႔ ေခၚပါတယ္) ရလာပါတယ္။ $ \displaystyle AB=BC=x$ ျဖစ္တာ ေၾကာင့္ $ \displaystyle AC$ ရဲ့ အလ်ားကို Pythagoras' Theorem နဲ႔ တြက္ယူႏိုင္ပါတယ္။

 Pythagoras' Theorem အရ

        $ \displaystyle \begin{array}{l}\ \ \ \ A{{C}^{2}}=A{{B}^{2}}+B{{C}^{2}}\\\\\ \ \ \ \ \ \ \ \ \ \ ={{x}^{2}}+{{x}^{2}}\\\\\ \ \ \ \ \ \ \ \ =2{{x}^{2}}\\\\\therefore \ \ AC=\sqrt{2}x\end{array}$

ဒါဆိုရင္ $ \displaystyle AB:BC:AC=1:1:\sqrt{2}$  ဆိုၿပီး အနားအခ်ိဳးေတြကို အလြယ္တကူ သိႏိုင္ပါၿပီ။ လက္ေတြ႕ တိုင္းတာမႈ မလုပ္ပဲ သခၤ်ာရဲ့ မွန္ကန္ခ်က္မ်ားျဖင့္ အနားအခ်ိဳးေတြကို အလြယ္တကူရွာႏိုင္တဲ့ $ \displaystyle 45°-45°$ right triangle ကို special triangle လို႔ ေခၚၿပီး $ \displaystyle 45°$ ေထာင့္ကိုေတာ့ special angle လို႔ ေခၚပါတယ္။

အနားေတြရဲ့ အခ်ိဳးကို သိၿပီဆိုေတာ့ $ \displaystyle 45°$ angle ရဲ့ trigonometric ratios ေတြကို အလြယ္ တကူရွာႏိုင္ၿပီေပါ့။

Trigonometric Ratios of $ \displaystyle 45°$

$ \displaystyle \sin 45{}^\circ =\frac{x}{{\sqrt{2}x}}=\frac{1}{{\sqrt{2}}}=\frac{{\sqrt{2}}}{2}$

$ \displaystyle \cos 45{}^\circ =\frac{x}{{\sqrt{2}x}}=\frac{1}{{\sqrt{2}}}=\frac{{\sqrt{2}}}{2}$

$ \displaystyle \tan 45{}^\circ =\frac{x}{x}=1$

$ \displaystyle \cot 45{}^\circ =\frac{x}{x}=1$

$ \displaystyle \sec 45{}^\circ =\frac{{\sqrt{2}x}}{x}=\sqrt{2}$

$ \displaystyle \operatorname{cosec}45{}^\circ =\frac{{\sqrt{2}x}}{x}=\sqrt{2}$

ယခုတစ္ခါ အနားတစ္ဘက္ $ \displaystyle 2x$ unit ရွိတဲ့ သံုးနားညီႀတိဂံတစ္ခု $ \displaystyle ABD$ ကို စဥ္းစားၾကည့္မယ္။ ေထာင့္စြန္းမွတ္ $ \displaystyle A$ မွ $ \displaystyle BD$ ေပၚသို႔ အျမင့္မ်ဥ္း $ \displaystyle AC$ ကိုဆြဲလိုက္မယ္ ဆိုရင္ သံုးနားညီ ႀတိဂံျဖစ္တာေၾကာင့္ $ \displaystyle AC$ ဟာ အလယ္မ်ဥ္းလည္း ျဖစ္သလို ေထာင့္ထက္၀က္ပိုင္း မ်ဥ္းလဲ ျဖစ္ပါတယ္။ ဒါ့ေၾကာင့္ ပံုမွာ ျပထားတဲ့ အတိုင္း $ \displaystyle AC$ ဟာ $ \displaystyle \vartriangle ABD$ ကို ထပ္တူညီ ႀတိဂံ ႏွစ္ခုအျဖစ္ ပိုင္းျဖတ္ လိုက္ပါတယ္။

၎တို႔အထဲက ေထာင့္မွန္ႀတိဂံ $ \displaystyle ABC$ ကို ခြဲထုတ္လိုက္ရင္ $ \displaystyle \vartriangle ABC$ ဟာ $ \displaystyle 30°-60°$ right triangle ျဖစ္ၿပီး $ \displaystyle AC$ ရဲ့ အလ်ားကိုေတာ့ Pythagoras' Theorem နဲ႔ တြက္ယူႏိုင္ပါတယ္။

 Pythagoras' Theorem အရ

$ \displaystyle \begin{array}{l}\ \ \ \ A{{C}^{2}}=A{{B}^{2}}-B{{C}^{2}}\\\\\ \ \ \ \ \ \ \ \ \ \ =4{{x}^{2}}-{{x}^{2}}\\\\\ \ \ \ \ \ \ \ \ \ \ =3{{x}^{2}}\\\\\therefore \ \ \ \ AC=\sqrt{3}x\end{array}$

ဒါဆိုရင္ $ \displaystyle BC:AB:AC=1:2:\sqrt{3}$  ဆိုၿပီး အနားအခ်ိဳးေတြကို အလြယ္တကူ သိႏိုင္ပါၿပီ။ လက္ေတြ႕ တိုင္းတာမႈ မလုပ္ပဲ သခၤ်ာရဲ့ မွန္ကန္ခ်က္မ်ားျဖင့္ အနားအခ်ိဳးေတြကို အလြယ္တကူရွာႏိုင္တဲ့ $ \displaystyle 30°-60°$ right triangle ကိုလည္း special triangle လို႔ ေခၚၿပီး $ \displaystyle 30°$ နဲ႔ $ \displaystyle 60°$ ေထာင့္ ေတြကိုေတာ့ special angle လို႔ ေခၚပါတယ္။

Trigonometric Ratios of $ \displaystyle 30°$

$ \displaystyle \sin 30{}^\circ =\frac{x}{{2x}}=\frac{1}{2}$

$ \displaystyle \cos 30{}^\circ =\frac{{\sqrt{3}x}}{{2x}}=\frac{{\sqrt{3}}}{2}$

$ \displaystyle \tan 30{}^\circ =\frac{x}{{\sqrt{3}x}}=\frac{1}{{\sqrt{3}}}=\frac{{\sqrt{3}}}{3}$

$ \displaystyle \cot 30{}^\circ =\frac{{\sqrt{3}x}}{x}=\sqrt{3}$

$ \displaystyle \sec 30{}^\circ =\frac{{2x}}{{\sqrt{3}x}}=\frac{2}{{\sqrt{3}}}=\frac{{2\sqrt{3}}}{3}$

$ \displaystyle \operatorname{cosec}30{}^\circ =\frac{{2x}}{x}=2$

Trigonometric Ratios of $ \displaystyle 60°$

$ \displaystyle \sin 60{}^\circ = \frac{{\sqrt{3}x}}{{2x}}=\frac{{\sqrt{3}}}{2}$

$ \displaystyle \cos 60{}^\circ =\frac{x}{{2x}}=\frac{1}{2}$

$ \displaystyle \tan 60{}^\circ = \frac{{\sqrt{3}x}}{x}=\sqrt{3}$

$ \displaystyle \cot 60{}^\circ =\frac{x}{{\sqrt{3}x}}=\frac{1}{{\sqrt{3}}}=\frac{{\sqrt{3}}}{3}$

$ \displaystyle \sec 60{}^\circ = \frac{{2x}}{x}=2$

$ \displaystyle \operatorname{cosec}60{}^\circ =\frac{{2x}}{{\sqrt{3}x}}=\frac{2}{{\sqrt{3}}}=\frac{{2\sqrt{3}}}{3}$

Basic Acute Angle

❀       The acute angle between the terminal side and the X - axis is called the basic acute angle.

❀       The basic acute angle is a positive acute angle.

❀       X-axis ႏွင့္ terminal side ၾကားတြင္ ရွိေသာ ေထာင့္က်ဥ္းကို basic acute angle ဟု ေခၚသည္။

❀       basic acute angle ကို အေပါင္းေထာင့္ အျဖစ္ အၿမဲသတ္မွတ္သည္။

❀       ေအာက္ပါဥပမာမ်ားကို ၾကည့္ပါ။

သတ္မွတ္ထားေသာ (ေပးထားေသာ) ေထာင့္ကို principal angle ဟုေခၚသည္။

Principal Angle ၏ trigonometric ratios မ်ား ရွာရန္

➤       principal angle ကို coordinate system (cartesian plane) တြင္ ေနရာခ်ပါ။

➤       သက္ဆိုင္ေသာ basic acute angle ကို ရွာပါ။

➤       basic acute angle ၏ sin ratio ႏွင့္ cos ratio ကို ရွာပါ။ (sin ႏွင့္ cos ကို သိလွ်င္ က်န္ေသာအခ်ိဳးမ်ားကို အလြယ္တကူ သိႏိုင္ပါသည္။)

➤       principal angle ႏွင့္ ၎ႏွင့္သက္ဆိုင္ေသာ basic acute angle တို႔၏ trigonometric ratio မ်ားသည္ ကိန္းဂဏန္းပမာဏ အားျဖင့္ (numerically) တူညီၾကပါသည္။

➤       သို႔ေသာ္ principal angle ၏ trigonometric ratio မ်ားအတြက္ သက္ဆိုင္ရာ quadrant ႏွင့္ ညိႇ၍ လကၡဏာ သတ္မွတ္ေပးရပါမည္။

Example : Find the six trigonometric ratios of 120°.

$ \displaystyle \ \ \ \sin 120{}^\circ =\sin 60{}^\circ \ \text{and}\ \cos 120{}^\circ =\cos 60{}^\circ \ \text{numerically}\text{.}$

$ \displaystyle \ \ \ \text{But }120{}^\circ \ \text{lies in the second quadrand}\text{.}$

$ \displaystyle \therefore \ \sin 120{}^\circ =\sin 60{}^\circ =\frac{{\sqrt{3}}}{2}$

$ \displaystyle \ \ \ \cos 120{}^\circ =-\cos 60{}^\circ =-\frac{1}{2}$

$ \displaystyle \therefore \ \tan 120{}^\circ =-\sqrt{3}$

$ \displaystyle \ \ \ \cot 120{}^\circ =-\frac{{\sqrt{3}}}{3}$

$ \displaystyle \ \ \ \sec 120{}^\circ =-2$

$ \displaystyle \ \ \ \operatorname{cosec}120{}^\circ =-\frac{{2\sqrt{3}}}{3}$

Trigonometric Ratios of ( – θ)

$ \displaystyle \ \ \ \text{In}\ \vartriangle PON,$

$ \displaystyle \ \ \ \sin \theta =y$

$ \displaystyle \ \ \ \cos \theta =x$

$ \displaystyle \ \ \ \tan \theta =\frac{y}{x}$

$ \displaystyle \ \ \ \cot \theta =\frac{x}{y}$

$ \displaystyle \ \ \ \sec \theta =\frac{1}{x}$

$ \displaystyle \ \ \ \operatorname{cosec}\theta =\frac{1}{y}$

$ \displaystyle \ \ \ \text{Since}\ \vartriangle {P}'O{N}'\cong \vartriangle PON,$

$ \displaystyle \ \ \ {y}'=y\ \text{and }{x}'=x\ \text{numerically}\text{.}$

$ \displaystyle \begin{array}{l}\ \ \ \text{But it is clear that, and }{y}'\ \text{have }y\ \text{opposite signs}\\\ \ \text{ }{x}'\ \text{and }x\ \text{have the same sign,}\end{array}$

$ \displaystyle \therefore {y}'=-y\ \text{and }{x}'=x.$

$ \displaystyle \therefore \sin (-\theta )={y}'=-y=-\sin \theta $

$ \displaystyle \ \ \ \cos (-\theta )={x}'=x=\cos \theta $

$ \displaystyle \ \ \ \tan (-\theta )=\frac{{{y}'}}{{{x}'}}=-\frac{y}{x}=-\tan \theta $

$ \displaystyle \ \ \ \cot (-\theta )=\frac{{{y}'}}{{{x}'}}=-\frac{x}{y}=-\cot \theta $

$ \displaystyle \ \ \ \sec (-\theta )=\frac{1}{{{x}'}}=\frac{1}{x}=\sec \theta $

$ \displaystyle \ \ \ \operatorname{cosec}(-\theta )=\frac{1}{{{y}'}}=-\frac{1}{y}=-\operatorname{cosec}\theta $

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Trigonometric Ratios of (270° + θ)

$ \displaystyle \ \ \ \text{In}\ \vartriangle PON,$

$ \displaystyle \ \ \ \sin \theta =y$

$ \displaystyle \ \ \ \cos \theta =x$

$ \displaystyle \ \ \ \tan \theta =\frac{y}{x}$

$ \displaystyle \ \ \ \cot \theta =\frac{x}{y}$

$ \displaystyle \ \ \ \sec \theta =\frac{1}{x}$

$ \displaystyle \ \ \ \operatorname{cosec}\theta =\frac{1}{y}$

$ \displaystyle \ \ \ \text{Since}\ \vartriangle {P}'O{N}'\cong \vartriangle OPN,$

$ \displaystyle \ \ \ {y}'=x\ \text{and }{x}'=y\ \text{numerically}\text{.}$

$ \displaystyle \ \ \ \text{But }{P}'({x}',{y}')\ \text{lies in the fourth quadrant}\text{.}$

$ \displaystyle \therefore {y}'=-x\ \text{and }{x}'=y.$

$ \displaystyle \ \ \ \sin (270{}^\circ +\theta )={y}'=-x=-\cos \theta $

$ \displaystyle \ \ \ \cos (270{}^\circ +\theta )={x}'=y=\sin \theta $

$ \displaystyle \ \ \ \tan (270{}^\circ +\theta )=\frac{{{y}'}}{{{x}'}}=-\frac{{x}}{{y}}=\cot \theta $

$ \displaystyle \ \ \ \cot (270{}^\circ +\theta )=\frac{{{x}'}}{{{y}'}}=-\frac{{y}}{{x}}=\tan \theta $

$ \displaystyle \ \ \ \sec (270{}^\circ +\theta )=\frac{1}{{{x}'}}=\frac{1}{y}=\operatorname{cosec}\theta $

$ \displaystyle \ \ \ \operatorname{cosec}(270{}^\circ +\theta )=\frac{1}{{{y}'}}=-\frac{1}{x}=-\sec\theta $

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